I have an 3x 18650 headlamp with a LED array. I’m a little curious as to how LED arrays power on/off?
Went camping over the weekend. I was surprised that when I turned my headlamp off, one of the four diodes remained dimly illuminated sufficient enough that it still illuminates my face. Photo of the diodes below.
Hopefully this array is a lesser drain than leaving batteries in old style flashlights? I’ll leave a single battery in it for a day and see if the battery voltage drops meaningfully. Starting at 3.860V
That actually shouldn’t happen. Those arrays are just meant to be turned on/off all at once, depending how the chips are connected internally. “3V” means all 4 are in parallel (4P), “6V” means 2 strings of 2 are in parallel (2S2P), and “12V” means all 4 are in series (4S).
Only 1 on implies a 3V LED, some leakage current even when “off”, and that lit one has a lower forward voltage than the other 3.
Eg, drill 4 holes in the middle of a barrel of water, 3 even and 1 slightly lower. When full, all 4 will be spewing water, but with just enough water in it, that 1 low-hole will be dripping water, and the other 3 nothing.
Thank you for the great explanation!
After 24 hours this single battery only dropped to 3.859V so a 24 hour loss of 0.001V which is what I’d expect for a stored battery anyways.
I’ll assume the worst and that the leakage can and will increase as time goes on.
Despite the leaking diode, my mind is blown how energy efficient these cheap present day LEDs are with getting several hours of super bright light. Cheers!